MICROECONOMICS

Preference Theory

Preference

Order Theory

Definition: (Binary Relation) A Binary Relation on $X$ is a subset $R$ on $X\times X$, if $(x,y) \in R$, we write $xRy$ and $xRyRz$.

Definition: (Reflexivity) A relationship is reflexive if $xRx\in R$.

Definition: (Transitivity) A relationship is transitive if $xRyRz \in R$ implies $xRz\in R$.

Definition: (Completeness) A relationship is complete if either $xRy\in R$ or $yRx\in R$.

Definition: (Symmetry) A relationship is symmetric if $xRy\in R$ implies $yRx\in R$.

Definition: (Anti-Symmetry) A relationship is anti-symmetric if $xRyRx\in R$ and implies $y=x$.

Definition: (Preorder) $R$ is a Preorder if $R$ is reflexive and transitive.

Definition: (Partial Order) R is a Partial Order if $R$ is transitive and antisymmetric.

Definition: (Linear Order) R is a Linear Order if $R$ is a complete partial order.

Preference

Assumption: (Consumption) The consumer has preferences over consumption bundle $X$, which is a nonempty, convex, compact subset of $\R^n$. $0$ is always in $X$.

Definition: (Preference) A Preference relation is a binary relation on the set of alternatives $X$, denoted as $"\succsim"$.

Definition: (Strict Preference) A Strict Preference relation is a binary relation on the set of alternatives $X$, denoted as $"\succ"$, if $x\succsim y$ but $y\nsucceq x$.

Definition: (Indifferent Preference) An Indifferent Preference relation is a binary relation on the set of alternatives $X$, denoted as $"\succ"$, if $x\succsim y$ but $y\nsucceq x$.

Definition: (At Least As Good Set) At Least As Good Set is defined as $\succsim(x^0) = \{x\in X, x\succsim x^0\}$.

Definition: (No Better Than Set) No Better Than Set is defined as $\precsim(x^0) = \{x\in X, x\precsim x^0\}$.

Definition: (Strictly Preferred To Set) Strictly Preferred To Set is defined as $\succ(x^0) = \{x\in X, x\succ x^0\}$.

Definition: (Strictly Worse Set) Strictly Worse Set is defined as $\prec(x^0) = \{x\in X, x\prec x^0\}$.

Definition: (Indifferent Curve) Indifferent Curve is defined as $\sim(x^0) = \{x\in X, x\sim x^0\}$.

Axiom

Axiom: (A1 - Completeness) For any distinct bundles $x$ and $y$, either $x\succsim y$ or $y\succsim x$.

Axiom: (A2 - Transitivity) For any distinct bundles $x,y$ and $z$, $x\succsim y\succsim z$ implies $x\succsim z$.

Lemma: (Completeness and Transitivity) if $\succsim$ is complete and transitive, then $\succ$ and $\sim$ are also complete and transitive.

Definition: (Rationality) A preference is called rational if it satisfies completeness and transitivity.

Theorem: (Ranking) Given a finite set of bundles $A\subset \R^n_+$, if the preference relation satisfies A1 and A2, then bundles in A can be ranked in a list consistent with the preference.

Proof:

It suffices to show that there is a maximum of $A$. Define an algorithm to find this item. Start with the first item in the set. Suppose next item is preferred to the item in hand, switch the items, i.e. if $x^m\succsim x^*$ but $x^m\nsucceq x^*$ do the switching. Keep doing this until we compare every item in the set.

The only potential trouble happens when we switch the items. We want to show that $x^*$ is preferred to any other item before the switching. Suppose there is no switching before. This implies that $x^1\succsim x^i$ for $i = 2,3,..., m-1$. From the switching we know that $x^m \succsim x^1$, by Transitivity, we have $x^* = x^m \succsim x^i$ for $i<m$. Now suppose there are $k$ numbers of switching happened before. We have $x^{m_k} \succsim x^i$ for $i$ between $m_{k-1}$ and $m_k$, plus we have $x^{m_k} \succsim x^{m_{k-1}}$. This implies that $x^* = x^m \succsim x^i$ for any $i<m$. If we repeat this algorithm, we will have $x^*\succsim x^i$ for any $i \in X$, i.e. $x^*$ is the maximum. $\square$

Axiom: (A3 - Continuity) For any bundle $x$, the set $\succsim(x)$ and $\precsim(x)$ are closed, or the set $\succ(x)$ and $\prec(x)$ are open.

Theorem: (Continuity) Suppose a preference is complete, transitive and continuous. Then for any $x\in X$, $y\in \succ(x)$ and $z\in \prec(x)$, there is a number $t\in (0,1)$ such that $ty+(1-t)z \in \sim(x)$.

Proof:

Define a subset of real number: $T = \{t\in (0,1)|ty+(1-t)z \in \succsim(x)\}$. Defined $t_0 = inf(T)$. Claim that $t_0 \in \sim(x)$. Suppose $t_0\nsim x$, then either $t_0\prec x$ or $t_0\succ x$. Suppose $t_0\prec x$, then since $\prec(x)$ is open, there is a $t>t_0$ such that $ty+(1-t)z \in \prec(x)$, contradicting to the fact that $t_0$ is the infimum. Suppose $t_0\succ x$, then since $\succ(x)$ is open, there is a $t<t_0$ such that $ty+(1-t)z \in \succ(x)$, again contradicting to the fact that $t_0$ is the infimum. $\square$

Axiom: (A4’ - Local Non-Satiation) For any $x$, for any $\epsilon>0$, there is another bundle $y\in B_\epsilon(x)$ such that $y\succ x$.

Axiom: (A4 - Strict Monotonicity) For any $x$ and $y$, if $y\geq x$ then $y\succsim x$, and if if $y\gg x$ then $y\succ x$.

Claim: (Local Non-Satiation and Strictly Monotonicity) If a preference is strictly monotonic then it is locally non-satiated.

Axiom: (A5’ - Convexity) If $y\succsim x$, then $ty+(1-t)x\succsim x$ for any $t\in [0,1]$.

Axiom: (A5 - Strict Convexity) If $y\succsim x$, and $y\neq x$, then $ty+(1-t)x\succ x$ for any $t\in (0,1)$.

Axiom: (A6 - Homothetic) If $y\sim x$, then $ty\sim tx$ for any $t>0$.

Utility Function

Existence of Utility Functions

Definition: (Utility Function) A function $U:\R^n_+\to \R$ is called a utility function representing the preference if for any $x,y\in X$, $U(x)\geq U(y)$ if and only if $x\succsim y$.

Theorem: (Existence of Utility Function) If the preference is complete, transitive, continuous, and strictly monatomic, then there is a real-valued and continuous function representing the preference.

Proof:

Let $e = [1,1,...,1]$ be a vector in $\R^n$. For any $x\in X$, define $U(x)$ so that $x\sim U(x)e$. We claim that such a number is well defined.

1. First we show that this number exists. Consider the following two sets: $A = \{t\geq 0|te\succsim x\}$ and $B = \{t\leq 0|te\precsim x\}$. Want to show that $A\cap B \neq \phi$. By continuity, both sets are closed. By strict monotonicity and transitivity, if $t_0\in A$, then $\forall t>t_0$ we have $t\in A$. Similarly, if $t_0\in B$, then $\forall t<t_0$ we have $t\in B$. So $A = [t_1,+\infty]$, and $B = [0,t_2]$. Suppose $t_1>t_2$, this would violate the completeness of the preference. Hence $t_1\leq t_2$, this number always exists.
2. Second we show that such a number is unique. Suppose $t_0<t_1$ and $x\sim t_0e\sim t_1 e$, by strict monatomicity, we have $t_0e\prec t_1 e$, which is a contradiction.

Now we will show that $U(x)$ represents the preference, i.e. $x\succsim y$ if and only if $U(x)\geq U(y)$. We have $U(x)e\sim x\succsim y\sim U(y)e$, and the statement is true as a result of strictly monotonicity.

The last step is to show $U(x)$ is continuous. It suffices to show that $U^{-1}(a,b)$ is an open set. $U^{-1}(a,b) = \{x\in X|a<U(x)<b\} = \{x\in X|ae \prec U(x)e \prec be\} = \{x\in X|ae \prec x \prec be\} = \succ(ae) \cap \prec(be)$. By continuity, we have both sets are opens, then the intersection of them is also open. $\square$

Definition: (Marginal Utility) The marginal utility is defined as $MU_i = \partial U/\partial x_i$.

Definition: (Marginal Rate of Substitution) The Marginal Rate of Substitution is defined as $MRS_{ij} = MU_i/MU_j$.

Definition: (Elasticity of Substitution) The Elasticity of Substitution is defined as $\sigma_{ij} =\frac{dln(x_i/x_j)}{dln(U_j/U_i)}$.

Properties

Lemma: (Utility and Preference) $U(x)> U(y)$ if and only if $x\succ y$ and $U(x) = U(y)$ if and only if $x\sim y$.

Theorem: (Monotonic Transformation) If $U$ is a function that represents the preference, and $V$ is another function representing the same preference, then $V = g(U)$ for some strictly increasing function $g:\R\to \R$. Inversely, if $V = g(U)$ for some strictly increasing function $g:\R\to \R$, then $V$ is another function representing the same preference.

Proof:

$U(x) \geq U(y)$ is equivalent to $g(U(x)) \geq g(U(y))$. Hence the statement is true. $\square$

Theorem: (Properties of Utility Function) The following statements are true:

1. $U$ is strictly increasing if and only if the preference is strictly monotonic
2. $U$ is quasi-concave if and only if the preference is convex
3. U is strictly quasi-concave if and only if the preference is strictly convex.

Proof:

These statements are true by definitions. Proof is omitted. $\square$

Definition: (Homogeneity) A utility function is called to be homogenous of degree $k$ if $U(tx) = t^kU(x) \forall t>0$.

Theorem: (Homogeneity and Homothetic Preference) A utility function is homogenous if and only if the preference is homothetic.

Proof:

Suppose a utility function is homogenous, then $U(tx) = t^kU(x)$. We have $x\sim y$ if and only if $U(x) = U(y)$. Then $U(tx) = t^kU(x) = t^kU(y) = U(ty)$, i.e. $tx\sim ty$, the preference is homothetic.

Suppose the preference is homothetic, then $x\sim y$ implies $tx\sim ty$. Define a utility function as $x\sim U(x)e$, where $e = (1,1,...,1)$. Then we have $tx \sim U(tx)e$. But homothetic preference implies that $tx \sim tU(x)e$. By transitivity we have $tU(x) = U(tx)$, i.e. $U$ is homogenous of degree 1. $\square$

Consumer Theory

Utility Maximization

Utility Maximization Problem

Assumption: (Consumer) The utility function is continuous, strictly increasing, strictly quasi-concave.

Definition: (Budget Set) Denote $B$ as the set of feasible choices for the consumer, it is a budget set when $B = \{x\geq0|px\leq I\}$. Note that there can be $k$ goods.

Definition: (Consumer’s Problem) The consumer is solving the Utility Maximizing Problem:

Theorem: (Existence and Uniqueness of the Solution) The solution to the consumer’s problem exists and is unique.

Proof:

The utility function is continuous, the budget set is compact, hence there is a maximum on the constraint.

Now we prove the uniqueness. Suppose there are two solutions to the problem, $x^1$ and $x^2$. Then $px^1\leq I$ and $px^2\leq I$, which implies that $px^t = p(tx^1+(1-t)x^2)\leq I$. Since both are solutions, we have $U(x^1) = U(x^2) = U$. Then $U(x^t)\leq U$, contradicting to the fact that $U$ is strictly quasi-concave. $\square$

Theorem: (Exhausting the Budget) With out the loss of generality, we could impose the budget constraint with equality.

Proof:

Suppose not, then the solution to the problem satisfies $px<I$, which cannot be true when $U$ is strictly increasing. $\square$

Solution

Solution: (Kuhn-Tucker Method)

Define the Lagrange Function as $L = U(x)+\lambda (I-px)+\mu x$. The Kuhn-Tucker condition to solve the problem is the first order conditions, i.e.

Theorem: (Kuhn-Tucker Theorem) Suppose $U$ is strictly quasi-concave and differentiable, $(p,I)\gg 0$, then if $(x^*,\lambda^*)\gg 0$ solve the first order conditions, then $(x^*,\lambda^*)$ solve the consumer’s problem.

Proof:

First notice one fact: when $x^1 \neq x$, $\nabla U(x)\neq 0$, and $U(x^1)>U(x)$, then $\nabla U(x)(x^1-x)>0$.

Now prove the statement. Suppose not, the there is another solution $(x^1,\lambda^1)$ such that $U(x^1)>U(x^*)$. By the fact above, we have $\nabla U(x^*)(x^1-x^*)>0$, and $px^1\leq I$. Notice that since $(x^*,\lambda^*)$ solve the Kuhn-Tucker conditions, we have $\nabla U(x^*) = \lambda^*p$. By assumption, $\lambda^*p\gg 0$, so $\nabla U(x^*)\gg 0$. So $\nabla U(x^*)(x^1-x^*) =\lambda^*p(x^1-x^*) >0$ implies that $px^1>px^* = I$, which is a contradiction to the fact that $x^1$ is feasible. $\square$

Definition: (Marshallian Demand) The solution the the utility maximizing problem is called the Marshallian Demand, denoted by $x^*(p,I)$.

Definition: (Elasticity) The Price Elasticity of the demand is defined as $\epsilon_i = \partial lnx^*(p,I)/\partial lnp_i$.

Indirect Utility Function

Definition: (Indirect Utility Function) The Indirect Utility Function, also known as the Value Function, is defined as $V(p,I) = U(x^*(p,I))$.

Theorem: (Properties of the Indirect Utility Function) If $U(.)$ is strictly increasing on $\R^n_+$, then $V(p,I)$ is:

1. Continuous on its domain
2. Homogenous of degree 0 in $(p,I)$
3. Strictly increasing in $I$
4. Decreasing in $p$
5. Quasi-convex in $(p,I)$
6. If $V(p,I)$ is differentiable and $\partial V/\partial I \neq 0$, then it satisfies Roy’s Identity: $-\frac{\partial V / \partial p_i}{\partial V/\partial I} = x_i(p,I)\geq 0$.

Proof:

1. The continuity of the value function follows from the theorem of maximum.

2. The value function is homogenous of degree 0 since $tpx\leq tI$ is identical to $px\leq I$.

3. consider $I'>I$, suppose $x(p,I)$ and $x'(p,I')$ are the solutions to the maximizing problem separately, then $px\leq I < I' = px'$. So consider $\epsilon \gg 0$ and $p(x+\epsilon)<I'$, but since the utility function is strictly increasing, $U(x+\epsilon)>U(x)$, hence $V(p,I')\geq U(x+\epsilon)>V(p,I)$.

4. consider $p'\leq p$, suppose $x(p,I)$ and $x'(p',I)$ are the solutions to the maximizing problem separately, then $p'x\leq px\leq I$ and $p'x'\leq I$. Then we have $V(p',I)\geq U(x) = V(p,I)$.

5. Now we want to show that $V$ is quasi-convex in $(p,I)$. It suffices to show that $B(p^1,I^1) \cup B(p^2,I^2) \supset B(\bar p,\bar I)$, where $B(\bar p,\bar I) = B(tp^1+(1-t)p^2,tI^1+(1-t)I^2)$. We will argue by contradiction. This means that there exists $\bar x \in B(\bar p, \bar I)$, such that $\bar x\notin B(p^1,I^1)$, and $\bar x\notin B(p^2,I^2)$. This implies that $p^1\bar x >I^1$ and $p^2\bar x >I^2$, then $tp^1\bar x >tI^1$ and $(1-t)p^2\bar x >(1-t)I^2$. Now if we combine them we would get $\bar p\bar x >\bar I$, contradicting to the fact that $\bar p\bar x \leq \bar I$.

6. The first proof follows from the Envelope Theorem. We have $\partial V / \partial p_i = \partial L/\partial p_i = -\lambda x_i$ and $\partial V / \partial I = \partial L/\partial I = \lambda$, which will give us the Roy’s Identity.

   Another proof defines $x^0 = x(p^0,I^0)$. Consider $V(p,px^0)-U(x^0)$ as a function of prices. Note that $V(p,px^0)-U(x^0)\geq 0$ for any $p\in \R^n_+$ because $x^0$ is always affordable. This function is minimized at the price $p^0$. Since the function is differentiable, we have the first order condition: $\partial V(p^0,p^0x^0)/\partial p_i+(\partial V(p^0,p^0x^0)/\partial I) x^0 = 0$, which will give us the Roy’s Identity. $\square$

Expenditure Minimization

Expenditure Minimizing Problem

Definition: (Expenditure Minimizing Problem) The expenditure minimizing problem is defined as

Theorem: (Existence of the solution) Let $U = \{u\in \R| \exist x \space s.t. \space U(x) = u\}$. If $U_0\in U$, and $U(.)$ is continuous, then there is a solution to the minimization problem. Furthermore, if $U(.)$ is strictly increasing and continuous and strictly quasi-concave, and $p\gg 0$, then the solution is unique.

Proof:

First, if $U_0\in U$, by definition there is a $x_0$ such that $U(x_0) = U_0$. So $px_0$ is enough to achieve the utility level. So the problem is equivalent to

Now we are minimizing a continuous function over a compact set, hence the solution exists.

Next we prove the uniqueness. Suppose there are two solutions to the problem, we have $p_x^1 = px^2 = I$ and $x^1\neq x^2$. Since $U(x^1)\geq U_0$ and $U(x^2)\geq U_0$, we have $U(x^t) = U(tx^1+(1-t)x^2)>min\{U(x^1),U(x^2)\} = U_0$ due to the strict quasi-concavity of the utility function. Note that $px^1 = px^2 = px^t$. Then there exists $x'\ll x^t$ such that $U(x')=U_0$ and $px'<I$, contradicting to the fact that $x_1$ and $x_2$ are both solutions. $\square$

Solution

Solution: (Lagrange Method)

Define the Lagrange function as $L = px+\lambda(U_0-U(x))$. Take the first order conditions we will get $p_i = \lambda\partial U/\partial x_i$ and $U(x) = U_0$.

Definition: (Hicks Demand) The solution the the expenditure minimizing problem is called the Hicks Demand, denoted by $x^H(p,U)$.

Expenditure Function

Definition: (Expenditure Function) The Expenditure Function is defined as $e(p,U) = px^H(p,U)$.

Theorem: (Properties of the Expenditure Function) If $U(.)$ is continuous, strictly increasing and strictly quasi-concave, then the expenditure function is:

1. Zero when $U$ takes the lowest utility level in available $U$
2. Continuous on $\R^N_+\times U$
3. For $p\gg 0$, the expenditure function is strictly increasing and unbounded in $U$
4. Increasing in $p$
5. Homogenous of degree 1 in $p$
6. concave in $p$
7. If $U(.)$ is differentiable, then the Shephard’s lemma is true, i.e. $\partial e(p^0,U^0)/\partial p_i = x_i^H(p^0,U^0)$.

Proof:

1. The bundle giving the lowest utility level is $(0,0,...,0)$. the expenditure to achieve this bundle is trivially zero.

2. Given the objective function is continuous and the feasible set is compact, by the Theorem of Maximum the expenditure function is continuous.

3. First prove the monotonicity. Suppose $U'>U$, let $px' = e(p,U')$, $U(x')\geq U'\geq U$, since $U(.)$ is strictly increasing and $px = e(p,U)$, we have $px = e(p,U)\leq e(p,U') = px'$, since $x$ is minimizing the objective function. Now we only need to show that $px'\neq px$. Suppose $px' = px$, since $U(x')\geq U'> U\geq U(0)$, we have $x'\neq 0$. There exists some $x_i'>0$ such that $U(\tilde x) = U(x_1',x_2',...,x_i'-\epsilon, ..., x_n')>U$ because $U'> U$ and $U$ is continuous. So $\tilde x$ is feasible for the minimization problem, and $px\leq p\tilde x<px'$, i.e. the expenditure function is strictly increasing in $U$.

   Now we prove the unboundedness. Since $U(.)$ is increasing and continuous, the attainable utility level is an interval: $U = [U(0),\bar U)$, where $\bar U$ is a finite number or infinity. Pick an increasing sequence of utility level $U_n\to \bar U$, we want to show that $e(p,U_n)$ is unbounded above. Suppose this is not true. Since $U(.)$ is strictly increasing, let $px^n = e(p,U_n)$, if $e(p,U_n)$ is bounded, then $x^n$ is also bounded. Then there is a subsequence of $x^n$,denoted as $x^{n_k}$ which goes to some $\bar x \in \R^n_+$. Then since $U(.)$ is continuous, we have $U(x^{n_k})\to U(\bar x)$. Since there is only one limit of the sequence $U_n$, which is $\bar U$, we have $U(\bar x) = \bar U$. But then $\bar U$ would be attainable, which is impossible.

4. Suppose $p^1<p^2$, and $x^1$ and $x^2$ are the solutions to the minimizing problem, separately. Since $U_0$ is always attainable, we have $p^1x^1\leq p^1x^2 < p^2x^2$.

5. At price $tp$, the problem can be rewrite as $t[min_{x\in \R^m_+} \space px] \space s.t. \space U(x)\geq U_0$, which is identical to the original problem. So the solution is the same, i.e. $e(tp,U_0) = tpx =t(px) = te(p,U_0)$.

6. Consider $p^1$ and $p^2$, define $\bar p = tp^1+(1-t)p^2$. Let $\bar x = x^H(\bar p, U_0)$, $x^1 = x^H(p^1, U_0)$, and $x^2 = x^H(p^2, U_0)$. Then we have $p^1x^1 \leq p^1 \bar x$ and $p^2x^2 \leq p^2 \bar x$. Combine them we have $\bar p (tx^1+(1-t)x^2)\leq \bar p\bar x$, i.e. the expenditure function is concave.

7. The first proof follows from the Envelope Theorem. We have $\partial e / \partial p_i = \partial L/\partial p_i = x_i^H$.

   An alternative proof is to consider the function $F(p) = e(p,U_0)-px^0$, where $x^0 = x^H(p,u^0)$. This function is maximized at $p$. So the first order condition is $\partial e(p^0,U^0)/\partial p_i - x_i^H(p^0,U^0) =0$. $\square$

Duality

Lemma: (Duality) Generally, we have $e(p,V(p,I))\leq I$, and $V(p,e(p,U))\geq U$.

Theorem: (Duality Theorem) If $U(.)$ is continuous and strictly increasing, then $e(p,V(p,I)) = I$ and $V(p,e(p,U))= U$.

Proof:

1. Prove by contradiction. Suppose $e(p,V(p,I_0)) < I_0$, let $U_0 =V(p,I_0)$ so then $e(p,U_0)<I_0$. Since $e(p,U)$ is continuous in $(p,U)$, there is a $\epsilon >0$ such that $e(p,U_0+\epsilon)<I_0$, denote $I_\epsilon = e(p,U_0+\epsilon)$, then $I_\epsilon<I_0$. But then we have $U_0+\epsilon \leq V(p,I_\epsilon)<V(p,I_0) = U_0$, which is impossible.
2. Prove by contradiction. Suppose $V(p,e(p,U_0)) > U_0$, let $I_0 = e(p,U_0)$ so then $V(p,I_0)>U_0$. Since $V(p,I)$ is continuous in $(p,I)$, there is a $\epsilon >0$ such that $V(p,I_0-\epsilon)>U_0$, denote $U_\epsilon = V(p,I_0-\epsilon)$, then $U_\epsilon>U_0$. But then we have $I_0-\epsilon \geq e(p,U_\epsilon)>e(p,U_0) = I_0$, which is impossible. $\square$

Theorem: (Duality of the Solutions) Suppose $U$ is continuous, strictly increasing and strictly quasi-concave. Then we have $x^* =x^H(p,V(p,I))$ and $x^H = x^*(p,e(p,U))$ for all $p\gg 0$ and U attainable.

Proof:

1. By continuity and strictly quasi-concavity, both solutions are unique. By definition we have $e(p,V(p,I)) = px^H(p,V(p,I))$. By the duality theorem $e(p,V(p,I)) =I = px^H(p,V(p,I))$. Then by definition of the utility maximizing problem, $x^H(p,V(p,I))$ is affordable under income level $I$ and gives a utility level of $V(p,I)$. So $x^H(p,V(p,I))$ is a solution to the utility maximizing problem. However, since the solution is unique, we have $x^* =x^H(p,V(p,I))$.
2. Similarly, by definition we have $V(p,e(p,U)) = U(x^*(p,e(p,U)))$. By the duality theorem $V(p,e(p,U)) = U = U(x^*(p,e(p,U)))$. Then by definition of the expenditure minimizing problem, $x^*(p,e(p,U))$ is attainable at utility level $U$ and gives an expenditure level of $e(p,U)$. So $x^*(p,e(p,U))$ is a solution to the expenditure minimizing problem. However, since the solution is unique, we have $x^H = x^*(p,e(p,U))$. $\square$